Question

Question 8 Three coins are tossed once. Find the probability of getting

(i) 3 heads (ii) 2 heads (iii) at least 2 heads

(iv) at most 2 heads (v) no head (vi) 3 tails

(vii) exactly two tails (viii) no tail (ix) at most two tails.

Class X1 - Maths -Probability Page 404

Answer 1

If three coins are tossed , then total number of possible outcomes = 2³ = 8
e.g { HHH,HHT, HTH, HTT,THH,THT,TTH,TTT}

(i) number of favourable outcomes n(E) = event getting 3 heads = 1 {e.g { HHH}
∵ P(E ) = n(E)/n(S)
= 1/8

(ii) number of getting 2 heads n(E) = 3 { {HTH}, {HTH} ,{THH}
P(E ) = n(E)/n(S) = 3/8

(iii) number of getting at least 2 heads n(E) = 4 { (HHH), (HHT), (HTH), (THH)}
P(E) = n(E)/n(S) = 4/8 = 1/2

(iv) number of getting almost 2 heads n(E) = 2 heads + 1 head + 0 head
= 3 + 3 + 1 = 7
P(E) = n(E)/n(S) = 7/8

(v) number of getting no head n(E) = 1
P(E) = n(E)/n(S) = 1/8

(vi) number of getting 3 tails n(E) = 1
P(E) = n(E)/n(S) = 1/8

(vii) number of getting 2 tails n(E) = 3
P(E) = n(E)/n(S) = 3/8

(viii) number of getting no tails n(E) = 1
P(E ) = n(E)/n(S) = 1/8

(ix) number of getting almost 2 tails n(E ) = 2 tails + 1 tails + 0 tail
= 3 + 3 + 1 = 7
P(E) = n(E)/n(S) = 7/8

Answer 2

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